Q. \[ \int e^{-x^{2}}\,dx \]

Q: What is the definite integral \int_{0}^{\infty} e^{-x^2}\,dx?

\int_{0}^{\infty} e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}.

Q: What is \int_{-\infty}^{\infty} e^{-x^2}\,dx?

\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}.

Q: How do you compute \int e^{-x^2}\,dx using series?

Expand e^{-x^2}=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{n!}, then integrate termwise: \int e^{-x^2}\,dx=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)n!}+C.

Q: What is the relation between \mathrm{erf}(x) and the integral \int e^{-x^2}\,dx?

By definition, \mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x} e^{-t^2}\,dt, so \int_{0}^{x} e^{-t^2}\,dt=\frac{\sqrt{\pi}}{2}\mathrm{erf}(x).

Q: Can we express \int e^{-ax^2}\,dx in terms of \mathrm{erf}?

Yes: \int e^{-a x^2}\,dx=\frac{\sqrt{\pi}}{2\sqrt{a}}\,\mathrm{erf}(\sqrt{a}\,x)+C for a>0.

Q: How do you differentiate \mathrm{erf}(x)?

\frac{d}{dx}\mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}e^{-x^2}.

Answer

We want the integral

\[
\int e^{-x^2}\,dx.
\]

This does not have an elementary antiderivative. Use the error function \(\operatorname{erf}(x)\), defined by

\[
\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x} e^{-t^2}\,dt.
\]

From this definition,

\[
\int e^{-x^2}\,dx=\frac{\sqrt{\pi}}{2}\operatorname{erf}(x)+C.
\]

Final result: \(\frac{\sqrt{\pi}}{2}\operatorname{erf}(x)+C\).

Detailed Explanation

We are asked to find the integral

\[
\int e^{-x^2}\,dx.
\]

Step 1: Recognize the standard difficulty.

The function \(e^{-x^2}\) does not have an elementary antiderivative (meaning it cannot be expressed using only polynomials, exponentials, logarithms, trig functions, etc.). So we use a special function.

Step 2: Use the definition of the error function.

The error function \(\operatorname{erf}(t)\) is defined by

\[
\operatorname{erf}(t)=\frac{2}{\sqrt{\pi}}\int_{0}^{t} e^{-u^2}\,du.
\]

Step 3: Match the integrand by a substitution.

We want an antiderivative of \(e^{-x^2}\). Compare

\[
\int_{0}^{x} e^{-u^2}\,du
\quad \text{with} \quad
\int_{0}^{t} e^{-u^2}\,du.
\]

If we set \(t=x\), then

\[
\int_{0}^{x} e^{-u^2}\,du=\frac{\sqrt{\pi}}{2}\operatorname{erf}(x).
\]

Step 4: Convert this definite-integral form into an indefinite integral.

An indefinite integral gives an antiderivative up to a constant. Since

\[
\int e^{-x^2}\,dx=\left(\int_{0}^{x} e^{-u^2}\,du\right)+C,
\]
we substitute the expression from Step 3:

\[
\int e^{-x^2}\,dx=\frac{\sqrt{\pi}}{2}\operatorname{erf}(x)+C.
\]

Final Answer.

\[
\int e^{-x^2}\,dx=\frac{\sqrt{\pi}}{2}\operatorname{erf}(x)+C.
\]

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Calculus FAQ

How do you evaluate the indefinite integral \(\int e^{-x^2}\,dx\)?

There is no elementary antiderivative. It is expressed using the error function: \(\int e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}\,\mathrm{erf}(x)+C\).

What is the definite integral \(\int_{0}^{\infty} e^{-x^2}\,dx\)?

\(\int_{0}^{\infty} e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}\).

What is \(\int_{-\infty}^{\infty} e^{-x^2}\,dx\)?

\(\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}\).

How do you compute \(\int e^{-x^2}\,dx\) using series?

Expand \(e^{-x^2}=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{n!}\), then integrate termwise: \(\int e^{-x^2}\,dx=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)n!}+C\).

What is the relation between \(\mathrm{erf}(x)\) and the integral \(\int e^{-x^2}\,dx\)?

By definition, \(\mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x} e^{-t^2}\,dt\), so \(\int_{0}^{x} e^{-t^2}\,dt=\frac{\sqrt{\pi}}{2}\mathrm{erf}(x)\).

Can we express \(\int e^{-ax^2}\,dx\) in terms of \(\mathrm{erf}\)?

Yes: \(\int e^{-a x^2}\,dx=\frac{\sqrt{\pi}}{2\sqrt{a}}\,\mathrm{erf}(\sqrt{a}\,x)+C\) for \(a>0\).

How do you differentiate \(\mathrm{erf}(x)\)?

\(\frac{d}{dx}\mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}e^{-x^2}\).

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Q. \[ \int e^{-x^{2}}\,dx \]

Answer

Detailed Explanation

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Calculus FAQ

How do you evaluate the indefinite integral \(\int e^{-x^2}\,dx\)?

What is the definite integral \(\int_{0}^{\infty} e^{-x^2}\,dx\)?

What is \(\int_{-\infty}^{\infty} e^{-x^2}\,dx\)?

How do you compute \(\int e^{-x^2}\,dx\) using series?

What is the relation between \(\mathrm{erf}(x)\) and the integral \(\int e^{-x^2}\,dx\)?

Can we express \(\int e^{-ax^2}\,dx\) in terms of \(\mathrm{erf}\)?

How do you differentiate \(\mathrm{erf}(x)\)?

Similar Questions

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